foundaons 1 lecture 5

Founda'ons1Lecture5

PatriciaA.Vargas

Lecture5

I.  MainPointsofLecture4II.  Proposi'onalLogic(Part5)

  Calcula'ngwithProposi'ons

2F29FAy‐Founda'ons1‐2009/2010‐Semester1‐PartI‐Logic&Proof

F29FAy‐Founda'ons1‐2009/2010‐Semester1‐PartI‐Logic&Proof 3

Commutativity, AssociativityIdempotence, double negation

Rules with True and FalseDistributivity, De Morgan

Rules with⇒Rules with⇔

Exercises

! val== has a lower priority than all connectives.

! Hence we should read Pval== Q ∧ R as P

val== (Q ∧ R).

!

Commutativity:

P ∧ Qval== Q ∧ P,

P ∨ Qval== Q ∨ P,

P ⇔ Qval== Q ⇔ P.

! P ⇒ Q does not have the same truth-value as Q ⇒ P.

P Q P ⇒ Q

0 0 10 1 11 0 01 1 1

P Q Q ⇒ P

0 0 10 1 01 0 11 1 1

Fairouz Kamareddine Logic and Proof F22HI1 Lecture 4




Exercises

! The Commutativity rules are easy to remember because ∧, ∨and ⇔ are their own mirror image on the vertical axis. (Thisis not the case though for the connective ⇒.)

! The word ’commutativity’ comes from the Latin word’commutare’, which is ’exchange’.

! Other very common cases of ’having the same truth-values as’or ’being equivalent to’ include:

Associativity:

(P ∧ Q) ∧ Rval== P ∧ (Q ∧ R),

(P ∨ Q) ∨ Rval== P ∨ (Q ∨ R),

(P ⇔ Q)⇔ Rval== P ⇔ (Q ⇔ R)

! Again, ⇒ is missing here.! The word ’associativity’ comes from the Latin word

’associare’, which means ’associate’, ’relate’.





Exercises

!Idempotence:

P ∧ Pval== P,

P ∨ Pval== P.

! The word ’idempotence’ comes from Latin. ’Idem’ means ’thesame’ and ’potentia’ means ’power’. It is actually the case forboth ∧ and ∨ that the ’second power’ of P is equal to P itself.

! You see this better when you replace ’∧’ by the multiplication

sign: P ∧ Pval== P becomes in that case P · P = P, and

hence P2 = P.

! In the same way you can write P ∨ Pval== P as P + P = P,

again a kind of second-power, but then for ’+’.





Exercises

!Double negation:

¬¬Pval== P.

! ’Double negation’ is also called ’involution’ after the Latin’involvere’, which is ’wrap’: when P becomes wrapped in two¬’s, it keeps its own truth value.

!Inversion:

¬True val== False,

¬False val== True.





Exercises

! The following two rules treat the confrontation of ¬ on onehand with ∧ resp. ∨ on the other.

! Also ¬ distributes in a certain sense, it ’exchanges ∧ and ∨’:

De Morgan:

¬(P ∧ Q)val== ¬P ∨ ¬Q,

¬(P ∨ Q)val== ¬P ∧ ¬Q





Exercises

! The standard equivalences so far took care of simplifications.! This will not always be the case with the rules of this section.! We begin with two rules which deal with a ’confrontation’

between ∧ and ∨:

Distributivity:

P ∧ (Q ∨ R)val== (P ∧ Q) ∨ (P ∧ R),

P ∨ (Q ∧ R)val== (P ∨ Q) ∧ (P ∨ R).

! The word ’distributivity’ comes from the Latin ’distribuere’,which is: ’distribute’ or ’divide’.

! In the first rule, for example, P and ∧ are together dividedover Q and R, because Q becomes P ∧ Q, and R becomesP ∧ R.

! One usually says by the first equivalence that ∧ distributesover ∨.





Exercises

!Double negation:

¬¬Pval== P.


!Inversion:

¬True val== False,

¬False val== True.





Exercises

! The rules below say what happens with True and False ifthey come across an ∧ or an ∨:

True/False-elimination:

P ∧ Trueval== P,

P ∧ Falseval== False,

P ∨ Trueval== True,

P ∨ Falseval== P.

! In the Boolean algebra where ∧ is written as ’.’ and ∨ iswritten as ’+’, True stands for the number 1 and False for0. The above True -False-laws become then:

P . 1 = P,P . 0 = 0,P + 1 = 1,P + 0 = P.





Exercises

! Another important rule which involves False, is the following:

Negation:

¬Pval== P ⇒ False

! Remember when reading this formula thatval== has the lowest

priority!! We already saw the equivalence of ¬P and P ⇒ False.! The above rule expresses that ’not-P’ ’means’ the same as ’P

implies falsity’.! Other important rules with True and False are:

Contradiction:

P ∧ ¬Pval== False

Excluded middle:

P ∨ ¬Pval== True.





Exercises


Negation:





Contradiction:


Excluded middle:






Exercises

! An important rule about ⇔:

Bi-implication:

P ⇔ Qval== (P ⇒ Q) ∧ (Q ⇒ P).

! A useful rule which follows as a consequence is the following:

Self-equivalence:

P ⇔ Pval== True.

! This can be seen as follows: First look at the formula P ⇒ P,the ’self-implication’, with unique arrow. It holds that:(1) P ⇒ P is equivalent to ¬P ∨ P, by Implication, and(2) ¬P ∨ P is in its turn equivalent to True, by (Commutativity

and) Excluded middle, therefore

(3) P ⇒ Pval== True.





Exercises

! There are many equivalences that can be constructed with’⇒’, we only give the most important ones.

! First we give the rules which change ⇒ into ∨, and vice versa.There, one always sees the appearance of ¬:

Implication:

P ⇒ Qval== ¬P ∨ Q,

P ∨ Qval== ¬P ⇒ Q.

! You can check these implications with truth tables.! Be careful that ¬ is always next to the first sub-formula.! If ¬ is next to the second sub-formula, then we lose the

equivalence:(1) P ∨ ¬Q is by Commutativity equivalent to ¬Q ∨ P, therefore

to Q ⇒ P, which is not equivalent to P ⇒ Q.(2) P ⇒ ¬Q is according to the first rule equivalent to ¬P ∨ ¬Q,

which is very different from P ∨ Q.





Exercises

! The following rule says something about the ’permutation’ ofan implication P ⇒ Q. Note the use of ¬.

Contraposition:

P ⇒ Qval== ¬Q ⇒ ¬P.

! We can check this with the truth tables. But we can alsocheck it as follows:(1) P ⇒ Q is equivalent to ¬P ∨ Q (by one Implication rule)(2) and this is equivalent to Q ∨ ¬P (by Commutativity),(3) which is equivalent to ¬Q ⇒ ¬P (by the otherImplication rule).





Exercises


Bi-implication:

P ⇔ Qval== (P ⇒ Q) ∧ (Q ⇒ P).


Self-equivalence:

P ⇔ Pval== True.





Proposi'onalLogic(Part4)InferenceRules(ExampleofCalcula'ngwithProposi'ons)





Exercises


Contraposition:

P ⇒ Qval== ¬Q ⇒ ¬P.



{Implica'onrule1}

{Implica'onrule2}{Commuta'vity}




Exercises


Contraposition:

P ⇒ Qval== ¬Q ⇒ ¬P.






Exercises



Implication:

P ⇒ Qval== ¬P ∨ Q,

P ∨ Qval== ¬P ⇒ Q.









Exercises



val== (Q ∧ R).

!

Commutativity:





P Q P ⇒ Q

0 0 10 1 11 0 01 1 1

P Q Q ⇒ P

0 0 10 1 01 0 11 1 1







Exercises


Bi-implication:

P ⇔ Qval== (P ⇒ Q) ∧ (Q ⇒ P).


Self-equivalence:

P ⇔ Pval== True.








Exercises



Implication:

P ⇒ Qval== ¬P ∨ Q,

P ∨ Qval== ¬P ⇒ Q.









Exercises



val== (Q ∧ R).

!

Commutativity:





P Q P ⇒ Q

0 0 10 1 11 0 01 1 1

P Q Q ⇒ P

0 0 10 1 01 0 11 1 1





Exercises


Negation:





Contradiction:


Excluded middle:






Exercises


Bi-implication:

P ⇔ Qval== (P ⇒ Q) ∧ (Q ⇒ P).


Self-equivalence:

P ⇔ Pval== True.








Exercises



P ∧ Trueval== P,



P ∨ Falseval== P.


P . 1 = P,P . 0 = 0,P + 1 = 1,P + 0 = P.







Exercises


Bi-implication:

P ⇔ Qval== (P ⇒ Q) ∧ (Q ⇒ P).


Self-equivalence:

P ⇔ Pval== True.





{Bi‐Implica'on}

{Commuta'vity,ExcludedMiddle}




Exercises


Bi-implication:

P ⇔ Qval== (P ⇒ Q) ∧ (Q ⇒ P).


Self-equivalence:

P ⇔ Pval== True.








Exercises


Bi-implication:

P ⇔ Qval== (P ⇒ Q) ∧ (Q ⇒ P).


Self-equivalence:

P ⇔ Pval== True.








Exercises


Bi-implication:

P ⇔ Qval== (P ⇒ Q) ∧ (Q ⇒ P).


Self-equivalence:

P ⇔ Pval== True.





(4) P⇔Pisequivalentto(P⇒P)∧(P⇒P)

(5) (P⇒P)∧(P⇒P)isequivalenttoTrue∧True {(3)}

(6) HencealsoequivalenttoTrue{True/False‐elimina'onrule1}

{Implica'onrule1}

Calcula'ngwithProposi'ons

Basic properties ofval==

Substitution, Leibniz’Calculations’ with equivalence

Equivalence in mathematicsExercises

! Lemma:(1) Reflexivity: P

val== P.

(2) Symmetry: If Pval== Q, then also Q

val== P.

(3) Transitivity: If Pval== Q and if Q

val== R, then P

val== R.

! Proof of part (1) The column of P in the truth-table is equalto the column of P (which is the same) in the truth-table.

! Proof of part (2) When the columns of P and Q in thetruth-table are equal, then so are those of Q and P.

! Proof of part (3) When in the truth-table, P and Q have thesame columns, and also Q and R have the same columns, thenthe columns of P and R must also be equal.








! Now we give an important relation between the

meta-symbolval== and the connective ⇔:

! Lemma: If Pval== Q, then P ⇔ Q is a tautology, and vice

versa.Proof:

! When Pval== Q, then P and Q have the same column of zeros

and ones in the truth-table. Hence, at the same height inthese columns, P and Q always have the same truth-value(either both 0, or both 1). In the column of P ⇔ Q there ishence always a 1, and so this is a tautology.

! On the other hand, when P ⇔ Q is a tautology then in thecolumn of P ⇔ Q there is everywhere a 1, hence at the sameheight in the columns of P and Q the same truth-values mustappear (either both 0, or both 1). Hence, these columns must

be completely equal, and so Pval== Q.







! For example let us take one of the two rules of De Morgan:

¬(P ∨ Q)val== ¬P ∧ ¬Q.

(Let Φ stand for (¬(P ∨ Q))⇔ (¬P ∧ ¬Q))

P Q P ∨ Q ¬(P ∨ Q) ¬P ¬Q ¬P ∧ ¬Q Φ

0 0 0 1 1 1 1 10 1 1 0 1 0 0 11 0 1 0 0 1 0 11 1 1 0 0 0 0 1

! From the equivalence of both columns under ¬(P ∨ Q) and

¬P ∧ ¬Q, it follows that ¬(P ∨ Q)val== ¬P ∧ ¬Q.

! The last column shows that (¬(P ∨ Q))⇔ (¬P ∧ ¬Q) is atautology.





! For example let us take one of the two rules of De Morgan:

¬(P ∨ Q)val== ¬P ∧ ¬Q.

(Let Φ stand for (¬(P ∨ Q))⇔ (¬P ∧ ¬Q))

P Q P ∨ Q ¬(P ∨ Q) ¬P ¬Q ¬P ∧ ¬Q Φ

0 0 0 1 1 1 1 10 1 1 0 1 0 0 11 0 1 0 0 1 0 11 1 1 0 0 0 0 1

! From the equivalence of both columns under ¬(P ∨ Q) and

¬P ∧ ¬Q, it follows that ¬(P ∨ Q)val== ¬P ∧ ¬Q.

! The last column shows that (¬(P ∨ Q))⇔ (¬P ∧ ¬Q) is atautology.







! The word ’substitution’ comes from the Latin ’substituere’,which means ’put in place’ or ’replace’.

! In mathematics and computer science substitution is thisprocess of repeatedly replacing one letter by the same formula.

! With simultaneous substitution more than one letter getsreplaced at the same time.

! Now it holds that equivalence is preserved with substitution:

! Lemma [Substitution] Suppose that a formula is equivalent toanother, and that, for example, the letter P occurs manytimes in both formulas. Then it holds that: If in bothformulas, we substitute something for P, then the resultingformulas are also equivalent. This holds for single, sequentialand simultaneous substitutions.





! The following lemma does not deal with substitution (for aletter P, Q, . . . ), but with a single replacement of asub-formula of a bigger formula.

! Lemma [Leibniz] If in an abstract proposition φ, sub-formulaψ1 is replaced by an equivalent formula ψ2, then the old andthe new φ are equivalent.

! We can write this as follows:ψ1

val== ψ2

. . . ψ1 . . .val== . . . ψ2 . . .

(the old φ) (the new φ)! In this picture you must read . . . ψ1 . . . and . . . ψ2 . . .

as being literally the same formulas, except for that one partwhere in the one we have ψ1 and in the other we have ψ2.







Exercises

! The symbolval== expresses the concept ’has the same

truth-value as ’ or ’is equivalent to’.

! val== is not a symbol of the logic itself, but a symbol from theso-called meta-logic .

! We use it exactly like a symbol in the logical formulas, but tospeak about the logical formulas.

! val== specifies a relation between logical formulas.

! Pval== Q is no abstract proposition itself.

! It only expresses that P and Q have always the sametruth-values: when one has truth-value 1 then the other willalso have truth-value 1, the same holds for the truth-value 0.

! ’val’ is a shorthand for ’value’.


P ∧ ¬Q ? ¬(P ⇒ Q )





Exercises



val== (Q ∧ R).

!

Commutativity:





P Q P ⇒ Q

0 0 10 1 11 0 01 1 1

P Q Q ⇒ P

0 0 10 1 01 0 11 1 1





Exercises




Associativity:

(P ∧ Q) ∧ Rval== P ∧ (Q ∧ R),

(P ∨ Q) ∨ Rval== P ∨ (Q ∨ R),

(P ⇔ Q)⇔ Rval== P ⇔ (Q ⇔ R)







Exercises

!Idempotence:

P ∧ Pval== P,

P ∨ Pval== P.




hence P2 = P.







Exercises

!Double negation:

¬¬Pval== P.


!Inversion:

¬True val== False,

¬False val== True.





Exercises



De Morgan:

¬(P ∧ Q)val== ¬P ∨ ¬Q,

¬(P ∨ Q)val== ¬P ∧ ¬Q





Exercises



Distributivity:

P ∧ (Q ∨ R)val== (P ∧ Q) ∨ (P ∧ R),

P ∨ (Q ∧ R)val== (P ∨ Q) ∧ (P ∨ R).








Exercises

!Double negation:

¬¬Pval== P.


!Inversion:

¬True val== False,

¬False val== True.





Exercises



P ∧ Trueval== P,



P ∨ Falseval== P.


P . 1 = P,P . 0 = 0,P + 1 = 1,P + 0 = P.





Exercises


Negation:





Contradiction:


Excluded middle:






Exercises


Negation:





Contradiction:


Excluded middle:






Exercises


Bi-implication:

P ⇔ Qval== (P ⇒ Q) ∧ (Q ⇒ P).


Self-equivalence:

P ⇔ Pval== True.








Exercises



Implication:

P ⇒ Qval== ¬P ∨ Q,

P ∨ Qval== ¬P ⇒ Q.









Exercises


Contraposition:

P ⇒ Qval== ¬Q ⇒ ¬P.






Exercises


Bi-implication:

P ⇔ Qval== (P ⇒ Q) ∧ (Q ⇒ P).


Self-equivalence:

P ⇔ Pval== True.










! Substitution and/or Leibniz are rarely named in the hint:

¬(P ⇒ Q)val== { Implication}

¬(¬P ∨ Q)val== { De Morgan}

¬¬P ∧ ¬Qval== { Double negation }

P ∧ ¬Q! The scheme above can have the following conclusion:

¬(P ⇒ Q)val== P ∧ ¬Q.

! Let us see another example where Leibniz, substitution,associativity and commutativity are used but not mentioned:




P ⇔ Q




Exercises

! The symbolval== expresses the concept ’has the same

truth-value as ’ or ’is equivalent to’.

! val== is not a symbol of the logic itself, but a symbol from theso-called meta-logic .

! We use it exactly like a symbol in the logical formulas, but tospeak about the logical formulas.

! val== specifies a relation between logical formulas.

! Pval== Q is no abstract proposition itself.

! It only expresses that P and Q have always the sametruth-values: when one has truth-value 1 then the other willalso have truth-value 1, the same holds for the truth-value 0.

! ’val’ is a shorthand for ’value’.


(P ∧ Q ) ∨ (¬P ∧ ¬Q ) ?





Exercises



val== (Q ∧ R).

!

Commutativity:





P Q P ⇒ Q

0 0 10 1 11 0 01 1 1

P Q Q ⇒ P

0 0 10 1 01 0 11 1 1





Exercises




Associativity:

(P ∧ Q) ∧ Rval== P ∧ (Q ∧ R),

(P ∨ Q) ∨ Rval== P ∨ (Q ∨ R),

(P ⇔ Q)⇔ Rval== P ⇔ (Q ⇔ R)







Exercises

!Idempotence:

P ∧ Pval== P,

P ∨ Pval== P.




hence P2 = P.







Exercises

!Double negation:

¬¬Pval== P.


!Inversion:

¬True val== False,

¬False val== True.





Exercises



De Morgan:

¬(P ∧ Q)val== ¬P ∨ ¬Q,

¬(P ∨ Q)val== ¬P ∧ ¬Q





Exercises



Distributivity:

P ∧ (Q ∨ R)val== (P ∧ Q) ∨ (P ∧ R),

P ∨ (Q ∧ R)val== (P ∨ Q) ∧ (P ∨ R).








Exercises

!Double negation:

¬¬Pval== P.


!Inversion:

¬True val== False,

¬False val== True.





Exercises



P ∧ Trueval== P,



P ∨ Falseval== P.


P . 1 = P,P . 0 = 0,P + 1 = 1,P + 0 = P.





Exercises


Negation:





Contradiction:


Excluded middle:






Exercises


Negation:





Contradiction:


Excluded middle:






Exercises


Bi-implication:

P ⇔ Qval== (P ⇒ Q) ∧ (Q ⇒ P).


Self-equivalence:

P ⇔ Pval== True.








Exercises



Implication:

P ⇒ Qval== ¬P ∨ Q,

P ∨ Qval== ¬P ⇒ Q.









Exercises


Contraposition:

P ⇒ Qval== ¬Q ⇒ ¬P.






Exercises


Bi-implication:

P ⇔ Qval== (P ⇒ Q) ∧ (Q ⇒ P).


Self-equivalence:

P ⇔ Pval== True.










!

P ⇔ Qval== { Bi-implication }

(P ⇒ Q) ∧ (Q ⇒ P)val== { Implication, twice }

(¬P ∨ Q) ∧ (¬Q ∨ P)val== { Distributivity }

(¬P ∧ (¬Q ∨ P)) ∨ (Q ∧ (¬Q ∨ P))val== { Distributivity, twice }

(¬P ∧ ¬Q) ∨ (¬P ∧ P) ∨ (Q ∧ ¬Q) ∨ (Q ∧ P)val== { Contradiction, twice }

(¬P ∧ ¬Q) ∨ False ∨ False ∨ (Q ∧ P)val== { True/False-elimination, twice }

(P ∧ Q) ∨ (¬P ∧ ¬Q)



Let’sworkonmoreexamples:•  Fortheproposi'onbelow,saywhetheritisatautology.

•  Ifyes,giveaproofbyacalcula'onsta'ngpreciselyateachstepwhichrulesofinferenceyouuse.

((Q⇒P)⇒¬Q)⇔(¬P∨¬Q)






Exercises



val== (Q ∧ R).

!

Commutativity:





P Q P ⇒ Q

0 0 10 1 11 0 01 1 1

P Q Q ⇒ P

0 0 10 1 01 0 11 1 1





Exercises




Associativity:

(P ∧ Q) ∧ Rval== P ∧ (Q ∧ R),

(P ∨ Q) ∨ Rval== P ∨ (Q ∨ R),

(P ⇔ Q)⇔ Rval== P ⇔ (Q ⇔ R)







Exercises

!Idempotence:

P ∧ Pval== P,

P ∨ Pval== P.




hence P2 = P.







Exercises

!Double negation:

¬¬Pval== P.


!Inversion:

¬True val== False,

¬False val== True.





Exercises



De Morgan:

¬(P ∧ Q)val== ¬P ∨ ¬Q,

¬(P ∨ Q)val== ¬P ∧ ¬Q





Exercises



Distributivity:

P ∧ (Q ∨ R)val== (P ∧ Q) ∨ (P ∧ R),

P ∨ (Q ∧ R)val== (P ∨ Q) ∧ (P ∨ R).








Exercises

!Double negation:

¬¬Pval== P.


!Inversion:

¬True val== False,

¬False val== True.





Exercises



P ∧ Trueval== P,



P ∨ Falseval== P.


P . 1 = P,P . 0 = 0,P + 1 = 1,P + 0 = P.





Exercises


Negation:





Contradiction:


Excluded middle:






Exercises


Negation:





Contradiction:


Excluded middle:






Exercises


Bi-implication:

P ⇔ Qval== (P ⇒ Q) ∧ (Q ⇒ P).


Self-equivalence:

P ⇔ Pval== True.








Exercises



Implication:

P ⇒ Qval== ¬P ∨ Q,

P ∨ Qval== ¬P ⇒ Q.









Exercises


Contraposition:

P ⇒ Qval== ¬Q ⇒ ¬P.






Exercises


Bi-implication:

P ⇔ Qval== (P ⇒ Q) ∧ (Q ⇒ P).


Self-equivalence:

P ⇔ Pval== True.





Lecture5I.  Proposi'onalLogic(Part5)  Calcula'ngwithProposi'ons

Material:

@hap://www.macs.hw.ac.uk/~pav1(pleasenotethatthislinkonlyworksinMozillaFirefox™orSafari®web

browsers)

@VISION


Lecture6What’snext?

PredicateLogicorFirst‐OrderLogic


foundaons 1 lecture 5

Documents